A map is continuous if and only if for every set, the image of closure is contained in the closure of image

As a part of self study, I am trying to prove the following statement:

Suppose X and Y are topological spaces and f:XY is a map. Then f is continuous if and only if f(¯A)¯f(A), where ¯A denotes the closure of an arbitrary set A.

Assuming f is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?


Because the property is stated in terms of closures, it’s I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:

If C is closed in Y, we need to show that D = f^{-1}[C] is closed in X.

Now using our closure property for D: f[\overline{D}] \subseteq \overline{f[D]} = \overline{f[f^{-1}[C]]} \subseteq \overline{C} = C, as C is closed.

This means that \overline{D} \subseteq f^{-1}[C] = D, making D closed, as required.

Source : Link , Question Author : Holdsworth88 , Answer Author : Henno Brandsma

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